Question

If the quadratic equation a(b-c)^2+b(cx)x+c(a+b)=0 has equal roots show that 1/a,1/b,1/c are in AP

Answer 1

Answer:

Step-by-step explanation:

Given quadratic equation,

a(b-c)x^2+b(c-a)^2+c(a-b)=0

also zeroes of given equation are equal.

Therefore,

discriminant=0

b^2-4ac=0

{b(c-a)}^2-4{a(b-c)}{c(a-b)}=0

b^2(c^2+a^2-2ac)-4{ab-ac}{ac-bc}=0

b^2c^2+a^2b^2-2ab^2c-4{a^2bc-ab^2c-a^2c^2+abc^2}=0

b^2c^2+a^2b^2-2ab^2c-4a^2bc+4ab^2c+4a^2c^2-4abc^2=0

b^2c^2+a^2b^2+2ab^2c-4a^2bc +4a^2c^2-4abc^2=0

b^2c^2+a^2b^2+4a^2c^2+2ab^2c -4a^2bc-4abc^2=0

We know that,

a^2+b^2+c^2+2ab+2bc+2ac={a+b+c}^2

By above identity we get,

{bc+ab-2ac}^2=0

bc+ab-2ac=0

b(a+c)=2ac

b=2ac/(a+c)

Hence if zeroes of given quadratic equation are equal then,

b=2ac/(a+c)